1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

I Why does a single sphere have a capacitance?

  1. Sep 14, 2016 #1

    The capacitance of an ideal plate capacitor (coaxial cable) goes to zero as the plate distance (outer radius) goes to infinity. This doesn't happen with concentric spheres as we let the outer radius go to infinity, hence a single sphere has a nonzero capacitance.

    What's the exact reason for this? Does it have something to do with the sphere being truly 3D while a plate is essentially 2D and a cable 1D?
    Or is this an artifact of the inconsistency in the derivation of the capacitances of the plate capacitor and coaxial cable? On one hand they assume a finite area/length, on the other they assume the electric field to behave as it would for infinite plates or cables.

    Different question: How does one compute the capacitance of an arbitrary conducting 3D shape? Is it always by first assuming an outer shell and then letting it go to infinity? What shape would this shell need to have?

    Yet another question: I assume a sphere has the maximum capacitance of some set of conducting 3D shapes, because it has no irregularities where the charge could accumulate and maximize the electric field. What is this set? Is it the set of all 3D shapes with the same volume, or the one of all 3D shapes with the same surface area, or something else?
  2. jcsd
  3. Sep 14, 2016 #2
    Look at the definition of capacitance. It is the charge that a conductor can hold divided by its potential (with respect to some reference). This definition makes no assumptions on shape. The problem becomes how do you determine the potential (difference) on a conductor given its charge. For the parallel plate capacitor the potential difference is given by the charge of the capacitor which depends on the voltage. the voltage determined by a battery is fixed but the charge depends on the separation. As the separation of the plates are increased using Gauss's the charge is decreased to maintain the voltage difference so as the separation becomes large the charge of the place decreases thus the capacitance decreases., For concentric spheres the charge on the inner sphere is fixed. and the potential difference is related to the spacing of the spheres. Since the potential ( wrt to infinity) on the inner sphere is fixed the outer sphere's separation as no effect on the inner sphere. If you took two parallel plates separated by a distance and placed a charge on one say with a electrostatic device.as you move the uncharged plate away the charge on the fixed plate would remain the same and have some potential and thus some capacitance. The problem would be to determine the potential of that plate .
  4. Sep 14, 2016 #3
    Actually every website and book I checked defines capacitance only in the context of capacitors, i.e. with two conductors and the relevant potential difference is the one between them. Is your (slightly more general) definition common?

    Isn't this some kind of circular argument? Aren't potential difference and voltage the same thing?

    Where do those two rather different constraints come in? All I see is two formulas for capacitance, both depending only on geometric properties, of which one gets zero and the other doesn't if certain (geometrically comparable) limits are taken. I don't see where either voltage in one case or charge in the other are fixed.
  5. Sep 14, 2016 #4
    Have you tried this one?

    Check the section "Self capacitance". And the section above it for a more general definition.
  6. Sep 14, 2016 #5
    Sorry my bad, it seems more like most websites and books motivate the concept of capacitance with capacitors and I misinterpreted this as a definition.

    What about my other questions?
  7. Sep 15, 2016 #6
    Voltage is a measure of potential difference.

    Capacitance of a conductor has one definition i.e. the ratio of the charge on a conductor divided by the potential of that conductor.

    C = Q/V

    When we talk of the potential of an object we are referring to the difference in potential of the object and infinity if it is a free standing conductor. The potential at infinity is taken as zero. Any free standing charged conductor has a capacitance.which means it can hold a charge.

    For a parallel plate capacitor we are looking at the capacitance between two conductors separated by a distance Δx. The charge arises in the capacitor because of the potential difference that was created by the battery. This potential difference is fixed. As the plates of the capacitor are moved farther apart the electric field decrease because E= ΔVfixed/Δx. But the electric field is also proportional to the charge density on the plate(s). So as the plates are separated more and more the electric field decreases and the charge on the capacitor plate(s) decreases. For a separation of infinity the charge on the plates goes to zero even though there is a potential difference.

    If you where to connect a battery between two concentric spheres as you increased the radius of the outside sphere you would see the capacitance of this configuration also go to zero.
  8. Sep 16, 2016 #7


    User Avatar
    Science Advisor
    Gold Member
    2017 Award

    Well, a charged sphere of radius ##R## leads to the electrostatic potential (in Heaviside-Lorentz units)
    $$V(\vec{r})=\frac{Q}{4 \pi r}, \quad r>0.$$
    The potential difference between a point at ##r=R## and ##r=R'>R## is
    $$\Delta V=\frac{Q}{4 \pi} \frac{R'-R}{R R'}.$$
    So your capacitance ##C=Q/\Delta V## depends on ##R## and ##R'## and thus this quantity only makes sense when you specify the reference point (in this case by giving ##R'##). Sometimes one defines the reference point for just a single sphere at ##R' \rightarrow \infty## and then the capacitance is given by ##C=4 \pi R## (in unrationalized Gaussian units simply ##R## for SI units ##4 \pi \epsilon_0 R##; all for vacuum).
  9. Sep 16, 2016 #8


    User Avatar
    Gold Member

    It one supposes that the "outer sphere" is at infinity then the concept seems meaningless. But in the practical case of the Hertzian Dipole, where spheres or plates are located on the ends of the two antenna rods, it will be found that the capacitance between the objects is much greater than the flat plate formula and corresponds roughly to the C=4 pi R concept (allowing for the two in series).
  10. Sep 18, 2016 #9
    Thanks everybody for their contributions. All your explanations sound logical to me, but I simply cannot follow them on a mathematical level. For a plate capacitor, we have
    $$C_1=\varepsilon\frac{A}{d}\enspace .$$
    For concentric spheres, we have
    $$C_2=\frac{4\pi\varepsilon}{\frac{1}{R_1}-\frac{1}{R_2}}\enspace .$$
    Obviously ##\lim_{d\to\infty} C_1=0## but ##\lim_{R_2\to\infty} C_2=4\pi\varepsilon R_1##. I just can't see where in this simple mathematical limit procedure all your assumptions about constant potential difference versus constant charge come in.

    How would you compute those limits with swapped assumptions? I.e. constant charge on the plate capacitor and constant potential difference between the two spheres.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted